Problem: Let $g(x)=\sqrt{x}\sin(x)$. $g'(x)=$
Answer: $g(x)$ is the product of two, more basic, expressions: $\sqrt{x}$ and $\sin(x)$. Therefore, the derivative of $g$ can be found using the product rule : $\begin{aligned} \dfrac{d}{dx}[u(x)v(x)]&=\dfrac{d}{dx}[u(x)]v(x)+u(x)\dfrac{d}{dx}[v(x)] \\\\ &=u'(x)v(x)+u(x)v'(x) \end{aligned}$ $\begin{aligned} &\phantom{=}g'(x) \\\\ &=\dfrac{d}{dx}(\sqrt{x}\sin(x)) \\\\ &=\dfrac{d}{dx}(\sqrt{x})\sin(x)+\sqrt{x}\dfrac{d}{dx}(\sin(x))&&\gray{\text{The product rule}} \\\\ &=\dfrac{d}{dx}\left(x^{^{\frac{1}{2}}}\right)\sin(x)+\sqrt{x}\dfrac{d}{dx}(\sin(x))&&\gray{\text{Write }\sqrt{x}\text{ as a power}} \\\\ &=\dfrac12x^{^{-\frac{1}{2}}}\cdot \sin(x)+\sqrt{x}\cdot \cos(x)&&\gray{\text{Differentiate }x^{^{\frac{1}{2}}}\text{ and }\sin(x)} \\\\ &=\dfrac{\sin(x)}{2\sqrt{x}}+\sqrt{x}\cos(x)&&\gray{\text{Simplify}} \end{aligned}$ In conclusion, $g'(x)=\dfrac{\sin(x)}{2\sqrt{x}}+\sqrt{x}\cos(x)$ or any other equivalent form.